Arrange data frame by all columns using dplyr

Question

I am generating data frames of 1s and 0s as follows:

library(tidyverse)
library(glue)

num_var <- 3

rep(list(c(0L, 1L)), num_var) %>%
  set_names(glue("var_{seq_len(num_var)}")) %>%
  expand.grid() %>%
  mutate(total = rowSums(.)) %>%
  select(total, everything()) %>%
  arrange(total, desc(var_1, var_2, var_3))

#>   total var_1 var_2 var_3
#> 1     0     0     0     0
#> 2     1     1     0     0
#> 3     1     0     1     0
#> 4     1     0     0     1
#> 5     2     1     1     0
#> 6     2     1     0     1
#> 7     2     0     1     1
#> 8     3     1     1     1

Created on 2018-01-08 by the reprex package (v0.1.1.9000).

I would need to arrange by the total sum of the variable in ascending order, and then each variable in descending order. This is fairly straightforward using dplyr::arrange(). However, I would like to have a more robust method of arranging. For example, if num_var is changed to, then, the final line must also be changed to arrange(total, desc(var_1, var_2, var_3, var_4)). I have tried using the tidy selector everything() to arrange as I do with the select() function, but this errors:

library(tidyverse)
library(glue)

num_var <- 3

rep(list(c(0L, 1L)), num_var) %>%
  set_names(glue("var_{seq_len(num_var)}")) %>%
  expand.grid() %>%
  mutate(total = rowSums(.)) %>%
  select(total, everything()) %>%
  arrange(total, desc(everything()))

#> Error in arrange_impl(.data, dots): Evaluation error: No tidyselect variables were registered.

Created on 2018-01-08 by the reprex package (v0.1.1.9000).

Is there a way to select variables for arranging without naming them all directly?

score 5 · Answer 1 · answered Feb 07 '22 at 17:34

For newer versions of dplyr, you can now use across:

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union

iris %>% 
  arrange(across(everything(), desc)) %>% 
  head()
#>   Sepal.Length Sepal.Width Petal.Length Petal.Width   Species
#> 1          7.9         3.8          6.4         2.0 virginica
#> 2          7.7         3.8          6.7         2.2 virginica
#> 3          7.7         3.0          6.1         2.3 virginica
#> 4          7.7         2.8          6.7         2.0 virginica
#> 5          7.7         2.6          6.9         2.3 virginica
#> 6          7.6         3.0          6.6         2.1 virginica


all.equal(iris %>% 
            arrange(across(everything(), desc)) ,
          iris %>% 
            arrange(desc(Sepal.Length), desc(Sepal.Width), desc(Petal.Length), desc(Petal.Width), desc(Species)))
#> [1] TRUE

^{Created on 2022-02-07 by the reprex package (v2.0.1)}

Psidom · Accepted Answer · 2018-01-08T16:04:42.933

arrange doesn't seem to work with select helper functions directly. You may use arrange_at, total in ascending order, and other variables except total (select using -one_of("total")) in descending order:

arrange_at(vars(total, desc(-one_of("total"))))

#  total var_1 var_2 var_3
#1     0     0     0     0
#2     1     1     0     0
#3     1     0     1     0
#4     1     0     0     1
#5     2     1     1     0
#6     2     1     0     1
#7     2     0     1     1
#8     3     1     1     1

score 1 · Answer 3 · answered Apr 08 '21 at 21:35

could arrange by every column going left to right using this code

library(magrittr) ; library(rlang) ; library(dplyr)
data %>% arrange(!!!syms(colnames(.)))

this works since arrange doesnt accept tidyselect syntax and thus must be passed symbols (or maybe also strings) for each of the names

Arrange data frame by all columns using dplyr

3 Answers3